Bernoulli equation: potential stream A stream is given by $(v, \rho, T)$.
The stress tensor is given by $\sigma = - p I$ with the pressure p, the force is given by $f = - \nabla \beta$. So $\beta$ is the potential of the force. 
If we have a potential stream, (i.e. $ v = \nabla \psi$ for a real function $\psi$), why this holds true? 
$$ \nabla ( \partial_t \psi + \frac{1}{2} |v|^2 + \beta) + \frac{1}{\rho} \nabla p = 0$$
I mean I know that $\nabla \times v = 0$, but I don't understand why the claim holds true...
 A: Start with the momentum equation for inviscid flow,
$$\tag{*}\frac{∂ \mathbf {v} }{∂ t}+\mathbf {v}\cdot \nabla \mathbf{v} = −\frac1ρ\nabla p+\mathbf{f} = −\frac1ρ∇p - \nabla \beta. $$
By Helmholtz' theorem, any continuously differentiable vector field that vanishes at infinity can be decomposed into  irrotational and soleniodal parts of the form
$$\mathbf{v} = \nabla \psi + \nabla \times \mathbf{a}.$$
For potential flow, we assume that the velocity field is irrotational, where $\nabla \times \mathbf{v} = 0$.  This requires that $\mathbf{a} = 0$ and, hence, 
$$\mathbf{v} = \nabla \psi.$$
We have the vector identity
$$=\nabla (|\mathbf{v}|^2) =\nabla(\mathbf{v} \cdot \mathbf{v}) = 2\mathbf{v}\cdot \nabla \mathbf{v} + 2\mathbf{v} \times (\nabla \times \mathbf{v}).$$
Since $\nabla \times \mathbf{v} = 0$ we get
$$\mathbf{v}\cdot \nabla \mathbf{v} = \nabla \left(\frac{1}{2} |\mathbf{v}|^2\right).$$
Substituting into (*) we get 
$$\frac{\partial}{\partial t} \nabla \psi +\nabla \left(\frac{1}{2} |\mathbf{v}|^2\right) = −\frac1ρ∇p - \nabla \beta,$$
and after interchanging $\frac{\partial}{\partial t} = \partial_t$ with $\nabla$ and rearranging,
$$\nabla \left( \partial_t \psi + \frac{1}{2} |\mathbf{v}|^2 + \beta\right) + \frac1\rho \nabla p = 0.$$
